Question

Angle α lies in quadrant II , and tanα=−125 . Angle β lies in quadrant IV , and cosβ=35 .

Answer 1

Answer:

$cos(\alpha+\beta)=\frac{33}{65}$

Step-by-step explanation:

step 1

Find cos α

we know that

$tan^2(\alpha)+1=sec^2(\alpha)$

we have

$tan(\alpha)=-\frac{12}{5}$

substitute

$(-\frac{12}{5})^2+1=sec^2(\alpha)$

$sec^2(\alpha)=\frac{144}{25}+1$

$sec^2(\alpha)=\frac{169}{25}$

$sec(\alpha)=\pm\frac{13}{5}$

Remember that Angle α lies in quadrant II

so

sec α is negative

$sec(\alpha)=-\frac{13}{5}$

Find the value of cos α

$cos)\alpha)=\frac{1}{sec(\alpha)}$

so

$cos(\alpha)=-\frac{5}{13}$

step 2

Find sin α

we know that

$tan(\alpha)=\frac{sin(\alpha)}{cos(\alpha)}$

$sin(\alpha)=tan(\alpha)cos(\alpha)$

we have

$tan(\alpha)=-\frac{12}{5}$

$cos(\alpha)=-\frac{5}{13}$

substitute

$sin(\alpha)=(-\frac{12}{5})(-\frac{5}{13})$

$sin(\alpha)=\frac{12}{13}$

step 3

Find sin β

we know that

$sin^2(\beta)+cos^2(\beta)=1$

we have

$cos(\beta)=\frac{3}{5}$

substitute

$sin^2(\beta)+(\frac{3}{5})^2=1$

$sin^2(\beta)=1-(\frac{3}{5})^2$

$sin^2(\beta)=1-\frac{9}{25}$

$sin^2(\beta)=\frac{16}{25}$

$sin(\beta)=\pm\frac{4}{5}$

Remember that

Angle β lies in quadrant IV

so

sin β is negative

$sin(\beta)=-\frac{4}{5}$

step 4

Find cos(α−β)

we know that

$cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)$

we have

$cos(\alpha)=-\frac{5}{13}$

$cos(\beta)=\frac{3}{5}$

$sin(\alpha)=\frac{12}{13}$

$sin(\beta)=-\frac{4}{5}$

substitute the given values

$cos(\alpha+\beta)=(-\frac{5}{13})(\frac{3}{5})-(\frac{12}{13})(-\frac{4}{5})$

$cos(\alpha+\beta)=(-\frac{15}{65})+(\frac{48}{65})$

$cos(\alpha+\beta)=\frac{33}{65}$